Exponents of numbers appear frequently in the calculation. We discuss both positive exponents and negative exponents. We show you how to handle such numbers perfectly.

### Definition of an integer exponent of a number

An exponent refers to the number of times a number is multiplied by itself. Let us for example take as number $5$. If $3$ is the exponent of $5$, then we obtain another number $5\times 5\times 5$ (the number five is repeated $3$ times if $3$ is the exponent of $5$). In this case we write $5^3$. Here $5$ is called the base, $3$ is the exponent and $5^3$ is the power.

Let us now give a more general formula for power numbers. Let $y$ be a number (natural of real number), and $n$ a natural number. We write

\begin{align*}{\Large y}^n={\Huge\underset{n \; {\rm times}}{\underbrace{y\times y\cdots\times y}}}\end{align*}

**Exercise:** let $a$ and $b$ be two numbers. Prove that

\begin{align*}(ab)^2=a^2b^2.\end{align*}

Solution: We know that $ab=ba$. Then by definition

\begin{align*}(ab)^2=(ab)(ab)=a(ba)b=a(ab)b=(aa)(bb)=a^2b^2.\end{align*}

More generally, if $n$ is a natural number, then

\begin{align*}

(ab)^n=a^nb^n.

\end{align*}

**Exercise: **Prove that

\begin{align*}

\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}

\end{align*}

Solution: We will apply the above formula of exponent of product numbers. In fact, we set $$c=\frac{a}{b}.$$

Then $bc=a$. Now we take the exponent in both sides of this equality, we get $(bc)^n=a^n$. But we know that $(bc)^n=b^nc^n$. Thus $b^nc^n=a^n.$ By dividing the both sides by $b^n,$ we obtain

\begin{align*}

\left(\frac{a}{b}\right)^n=c^n=\frac{a^n}{b^n}.

\end{align*}

Let $n$ and $m$ be two integers and $y$ be a real number. Then

\begin{align*}{\Large y}^{n+m}={\Large y}^n {\Large y}^m\end{align*}

** Application:** For intergers numbers $n$ and $m,$ prove that

\begin{align*} \left({\Large y}^{n}\right)^m={\Large y}^{nm}.\end{align*}

In fact, by definition of exponents, we have

\begin{align*} \left({\Large y}^{n}\right)^m&={\Large y}^n\cdot{\Large y}^n\cdots{\Large y}^n\;(m\; times)\cr & = {\Large y}^{n+n+\cdots+n}.\end{align*}

But the number $n$ is repeated $m$ times in the sum $n+n+\cdots+n$. Thus $n+n+\cdots+n=mn$. The result then follows.

Example: Compare the numbers $8^4$ and $2^{12}$.

\begin{align*}2^{12}= 2^{3\times 4}=(2^3)^4=8^4.\end{align*}

**Exercise:** Find the number $x$ such that $x^5=2^{15}$.

Solution: Let us first recall the following rule: $ a^n = b^n $ implies that $ a = b $. We will apply this property to answer the exercise. In fact, the idea is to rewrite $2^{15}$ as an exponent of $5$. Remark that $15=3\times 5$. Then

\begin{align*}x^5=2^{15}=2^{3\times 5}=(2^3)^5=8^5.\end{align*}

It follows that $x=8$.

### Negative exponents

In the previous section we have seen the definition and properties of postive exponents of numbers. In the actuel section we will discuss the case of negative exponenent of numbers. By definition if $a$ is a real number and $n$ is a integer number, then

\begin{align*}a^{-n}:=\frac{1}{a^n}.\end{align*}

For $n=1,$ we have

\begin{align*}a^{-1}=\frac{1}{a}.\end{align*}This shows that $$(a^{-1})^n=a^{-n}.$$

On the other hand, if $n$ and $m$ are integers such that $n\ge m$, then we can write $a^n=a^{(n-m)+m}=a^{n-m}a^m$. By dividing the two sides of this equation by $a^m$, we get

\begin{align*}a^{n-m}=a^n\times \frac{1}{a^m}=a^n a^{-m}.\end{align*}

From this formula we conclude that for any relative numbers (positive or negative) $n$ and $m$, we have

\begin{align*}{\Large y}^{n+m}={\Large y}^n {\Large y}^m\end{align*}

**Exercise: **Simplify the expression

\begin{align*}A(n)=\frac{2^n+1}{1+2^{-n}}.\end{align*}

Solution: We factor the numerator of the fraction by $2^n,$ we get

\begin{align*}A(n)=2^n\frac{1+2^{-n}}{1+2^{-n}}=2^n.\end{align*}

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